We can distinguish between these two models by doing what's called an epistasis test to establish the epistatic relationship between Gal4 and Gal80.This involves making a double Gal4/Gal80 mutant strain.The phenotype of the double mutant will indicate which of the two models is most likely to be true....take a look at the two models to predict what phenotype the double mutant should have.For Model 1 the double mutant would become uninducible,for Model 2 the double mutant should be constitutive We could make the Gal4/Gal80 double mutant strain using molecular biological approaches...but an easier way is to let yeast meiosis do the job for you.If we mate the Gal4/Gal80+haploid strain with the Gal4+/Gal80 haploid strain (we know these two genes are unlinked)we should obtain double mutants among the tetratype and non-parental ditype tetrads that result from this cross. Using double mutants to order Gal4 and Gal80 MATa Gal4-Gal80+x MATa Gal4+Gal80- Type 1 Type 2 Type 3 uninducible uninducible uninducible uninducible uninducible uninducible constitutive constitutive regulated (wt) constitutive regulated (wt) regulated (wt) Parental Non Parental Ditype Tetratype Ditype Gal4-Gal80+ Gal4-Gal80+ Gal4-Gal80- Gal4-Gal80+ Gal4-Gal80- Gal4-Gal80- Gal4+Gal80- Gal4+Gal80- Gal4+Gal80+ Gal4+Gal80- Gal4+Gal80+ Gal4+Gal80+ These results clearly support Model 1,i.e.,because the double mutant is uninducible rather than constitutive,Gal4 liklely behaves as a positive activator of Gal1 expression,and in the absence of galactose Gal80 somehow prevents Gal4 from activation Gal1 expression.When galactose is present Gal80 can no longer prevent Gal4 from activating Gal1 expression. Now lets consider a new class of mutant that turned out to be quite informative,Gal81.Gal81 mutants,like Gal80 mutants are consititutive for Gal1 expression,but unlike Gal80,Gal81 is dominant.(Gal81/Gal80 diploids are constitutive).We can distinguish between these two models by doing what’s called an epistasis test to establish the epistatic relationship between Gal4- and Gal80- . This involves making a double Gal4- / Gal80- mutant strain. The phenotype of the double mutant will indicate which of the two models is most likely to be true….take a look at the two models to predict what phenotype the double mutant should have. For Model 1 the double mutant would become uninducible, for Model 2 the double mutant should be constitutive. We could make the Gal4- / Gal80- double mutant strain using molecular biological approaches…but an easier way is to let yeast meiosis do the job for you. If we mate the Gal4- / Gal80+ haploid strain with the Gal4+ / Gal80- haploid strain (we know these two genes are unlinked) we should obtain double mutants among the tetratype and non-parental ditype tetrads that result from this cross. Parental Ditype Gal4- Gal80+ Gal4- Gal80+ Gal4+ Gal80- Gal4+ Gal80- Tetratype Gal4- Gal80+ Gal4- Gal80- Gal4+ Gal80- Gal4+ Gal80+ Non Parental Ditype Gal4- Gal80- Gal4- Gal80- Gal4+ Gal80+ Gal4+ Gal80+ Parental Ditype Gal4- Gal80+ Gal4- Gal80+ Gal4+ Gal80- Gal4+ Gal80- Tetratype Gal4- Gal80+ Gal4- Gal80- Gal4+ Gal80- Gal4+ Gal80+ Non Parental Ditype Gal4- Gal80- Gal4- Gal80- Gal4+ Gal80+ Gal4+ Gal80+ These results clearly support Model 1, i.e., because the double mutant is uninducible rather than constitutive, Gal4 liklely behaves as a positive activator of Gal1 expression, and in the absence of galactose Gal80 somehow prevents Gal4 from activation Gal1 expression. When galactose is present Gal80 can no longer prevent Gal4 from activating Gal1 expression. Now lets consider a new class of mutant that turned out to be quite informative, Gal81- . Gal81- mutants, like Gal80- mutants are consititutive for Gal1 expression, but unlike Gal80- , Gal81- is dominant. (Gal81- / Gal80- diploids are constitutive)